Chapter 5.3, Problem 83HP

To determine

To show: The axis of symmetry of quadratic equation $\left(x-p\right)\left(x-q\right)=0$ is halfway between the x -intercepts $p\text{ and }q$.

Answer to Problem 83HP

The axis of symmetry of quadratic equation $\left(x-p\right)\left(x-q\right)=0$ is $\frac{p+q}{2}$.

Explanation of Solution

Given information:

The quadratic equation $\left(x-p\right)\left(x-q\right)=0$.

Formula used:

The axis of symmetric of equation of the form $a{x}^2+bx+c=0$ is $x=\frac{-b}{2a}$.

Calculation:

Consider the quadratic equation $\left(x-p\right)\left(x-q\right)=0$.

Expand the above equation,

  $\begin{array}{c}\left(x-p\right)\left(x-q\right)=0\\ x^2-qx-px+pq=0\\ x^2+x\left(-p-q\right)+pq=0\end{array}$

Recall that the axis of symmetric of equation of the form $a{x}^2+bx+c=0$ is $x=\frac{-b}{2a}$.

Apply it, axis of symmetry of $x^2+x\left(-p-q\right)+pq=0$ is,

  $\begin{array}{c}x=-\frac{\left(-p-q\right)}{2}\\ x=\frac{p+q}{2}\end{array}$

It is known that midpoint of two numbers a and b is $\frac{a+b}{2}$.

So, axis of symmetry of quadratic equation $\left(x-p\right)\left(x-q\right)=0$ is halfway between the x -intercepts $p\text{ and }q$ which is equal to $x=\frac{p+q}{2}$.

Thus, the axis of symmetry of quadratic equation $\left(x-p\right)\left(x-q\right)=0$ is $\frac{p+q}{2}$.